Re: [EMHL] Fermat Points

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• Each of these triangles has a point U = midpoint of GH. The Fermat line goes through this point, so either the U points of the three triangles involved are
Message 1 of 11 , Dec 8, 2005
Each of these triangles has a point U = midpoint of GH. The Fermat
line goes through this point, so either the U points of the three
triangles involved are colinear, or they are the same.

Steve

On Dec 7, 2005, at 9:22 AM, Wilson Stothers wrote:

> Dear All,
>
> We have the following fact :
>
> If F is a Fermat Point of triangle ABC, then F is also a
> Fermat Point of its Cevian, and of its Anticevian Triangle.
>
> A couple of questions :
>
> Is this well-known? I'd appreciate a reference.
>
> Apart from G, are there other centres with this property?
>
> I also observe from Cabri that, if F is a Fermat point,
> and if ABC and the Cevian and Anticevian Triangles have
> Fermat Points {F,F2},{F,F3},{F,F4}, respectively, then
> (a) the points F,F2,F3,F4 are concyclic,
> (b) the circle also contains the isogonal conjugates of
> F2, F3, F4 with respect to their triangles,
> (c) the point F and its isogonal conjugates in the three
> triangles are collinear,
> (d) the Euler Lines of the three triangles are parallel,
> this is an easy consequence of (c).
>
> Of course the isogonal conjugates in (b) and (c) are the
> Isodynamic Points of their triangles. Thus, if F = X13,
> then the circle is that through X13, X14 and X16, and
> the line is X13X15. Note that F need not be X13 for the
> other two triangles, but is a Fermat Point.
>
> I suspect that (c) characterizes the Fermat Points, and
> perhaps that (d) also does.
>
> I'd be glad of any proofs or comments on these. I think
> that a circle with seven points is unusual. The centre
> might be interesting, it is on X115X125 as this is the
> perpendicular bisector of X13X14.
>
> Seasonal Greetings
>
> Wilson
>
>
>
>
>
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• Dear Bernard, I know nothing about angular coordinates, so hesitate to say much Your proof below suggests that, if F = X13, then it is also X13 of its cevian
Message 2 of 11 , Dec 8, 2005
Dear Bernard,

I know nothing about angular coordinates, so hesitate to say much
Your proof below suggests that, if F = X13, then it is also X13 of
its cevian and anticevian triangle. This isn't always so, it can
be X14 for one or other.

> > We have the following fact :
> >
> > If F is a Fermat Point of triangle ABC, then F is also a
> > Fermat Point of its Cevian, and of its Anticevian Triangle.
>
> this is direct consequence of the fact that the angular
coordinates
> of a point wrt ABC, its cevian and anticevian triangles are the
same.

Is this true?
I'm guessing at the definition, but I think there may be
supplements involved. Probably I just guessed wrongly.

> for X13 : 120°, 120°, 120°
>
> [snip]
>
>

Yours

Wilson
• Ah, that is what I wanted to know. But it seems to me that their angular coordinates would not change, so both points should be the same. It puzzles me that
Message 3 of 11 , Dec 8, 2005
Ah, that is what I wanted to know. But it seems to me that their
angular coordinates would not change, so both points should be the
same. It puzzles me that the second points are different. Intriguing
if they are.

Oh I know, it must be true that if we get the pedal and antipedal of
Fs, then the three Fs's are the same. This makes 12 points from these
six triangles, which I bet have an interesting relationship, Can't
wait until I have some time with Sketchpad.

Hmm, algebra at work.

Steve

On Dec 8, 2005, at 9:55 AM, Wilson Stothers wrote:

> the three triangles will have different second
> Fermat Points, so the lines FnFs are not parallel.

[Non-text portions of this message have been removed]
• Bernard, The two Fermat points each have a cevian and anticevian triangle. The 4 Euler lines of these triangles seem to be parallel to that of ABC (observed on
Message 4 of 11 , Dec 9, 2005
Bernard,

The two Fermat points each have a cevian and anticevian triangle. The
4 Euler lines of these triangles seem to be parallel to that of ABC
(observed on Sketchpad, not proved). Is there a curve, all of whose
points have Euler line parallel to that of ABC?

Steve
• Dear Steve, ... It is true. ... this was answered by Wilson yesterday under the thread [EMHL] The Cubic K278! Best regards Bernard [Non-text portions of this
Message 5 of 11 , Dec 9, 2005
Dear Steve,

> The two Fermat points each have a cevian and anticevian triangle. The
> 4 Euler lines of these triangles seem to be parallel to that of ABC
> (observed on Sketchpad, not proved).

It is true.

> Is there a curve, all of whose
> points have Euler line [of their cevian and anticevian triangles, I
> suppose] parallel to that of ABC?

Cubic K278!

Best regards

Bernard

[Non-text portions of this message have been removed]
• WILSON SOTHERS CIRCLE On Dec 7, 2005 Wilson Sothers posted to Hyacinthos (1): I also observe from Cabri that, if F is a Fermat point, and if ABC and the
Message 6 of 11 , Sep 6, 2016
WILSON SOTHERS CIRCLE

On Dec 7, 2005 Wilson Sothers posted to Hyacinthos (1):

<q> I also observe from Cabri that, if F is a Fermat point, and if ABC and the Cevian and Anticevian Triangles have
Fermat Points {F,F2},{F,F3},{F,F4}, respectively, then
(a) the points F,F2,F3,F4 are concyclic,
(b) the circle also contains the isogonal conjugates of F2, F3, F4 with respect to their triangles,
(c) the point F and its isogonal conjugates in the three triangles are collinear,
(d) the Euler Lines of the three triangles are parallel, this is an easy consequence of (c). </q>

On Aug 30, 2016, Tran Quang Hung posted (2):

<q>Let A'B'C' be cevian triangle of F1 and A''B''C'' be anticevian triangle of F1 wrt ABC.
F' and F'' are second Fermat points of triangles A'B'C' and A''B''C''.
Then four point F1,F2,F',F'' are concyclic on circle (w)
So if we construct more anticevian of F1 wrt A''B''C''.... then the second Fermat points of new triangles lie on (w), too.
Is this a new circle ? </q>

Note:
As the author of the page, says in the Foreword:
TTW started taking shape in the 2002 Spring originated from a wide collection of previous notes,  a great number of them borrowed from the Hyacinthos mail list and, after, of its follower Anopolis. Let me thank all its members.
http://www.xtec.cat/~qcastell/ttw/ttweng/forew1.html

C. Lozada posted the center of the circle (4):
<q>The center of the circle is X(9201)= X(14)-OF-2nd-PARRY-TRIANGLE. It passes through ETC's 13,14,15,16,5623 <q>

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